#f(x)=x^3-12 x +2#
#f^'(x)=3 x^2-12 #, critical points are those point where,
slope,#f'(x)=0 :. 3 x^2-12=0 or 3 (x^2-4)=0 #or
#3 (x+2)(x-2)=0 or x= -2 and x=2#
#f(-2)= (-2)^3-12*(-2)+2= 18 or (-2,18)# and
#f(2)= 2^3-12*2+2= -14 or (2, -14)#
Slope check in interval ,
# x< -2 , f^'(x)= (-)*(-)=(+) ; f'(x)>0 :. #
Therefore, increasing slope.
# -2< x<2 , f^'(x)= (+)*(-)=(-) ; f'(x)<0 :. #,
Therefore, decreasing slope .
# x> 2 , f^'(x)= (+)*(+)=(+) ; f'(x)>0 :. #
Therefore, increasing slope.
At #x=-2# the slope changes from increasing to decreasing,
so #(-2,18)# is local maximum point and
at #x=2# the slope changes from decreasing to increasing,
hence, #(2,-14)# is local minimum point.
graph{x^3-12x +2 [-40, 40, -20, 20]} [Ans]
